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3 years ago

What is the domain of a sine function?

2 answers:

7 0

Hello there!

The domain of the sine function is [-infinity, +infinity].

I hope this is the answer you were looking for. As always, it is my pleasure to help students like you!

Charra [1.4K]3 years ago

3 0

Answer: The domain of the sine function is (-∞, +∞).

Step-by-step explanation:

The domain of the function y = sin x is all real numbers (the sine is defined for any angle measure), and the range is −1 ≤ y ≤ 1.

So, the domain of the sine function is (-∞, +∞).

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a pet store has 2 kittens. one kitten weighs 15/16 pounds. the other kitten weighs 12/16 pounds. what is the difference of the w

Lostsunrise [7]

Subtract 15/16 and 12/16 to get 3/16

5 0

3 years ago

Jesse has two packages of 36TACKS and 16 TACKS. How many garage sale posters can she put up if she uses 4TACKS for each poster?

hodyreva [135]

Answer: 21 posters or 13 posters, please read the explanation since your wording was confusing.

Step-by-step explanation: Okay if you have two packs of 36 and 16 which is 104, you divide by 4 tacks. You end up getting 21 posters.

If you mean there are only 36 and 16 you get 52 and divide by 4. You end up having 13 posters.

7 0

3 years ago

I cant do it sombody help

motikmotik

Step-by-step explanation:

hope this helps. in the first part you set the problems equal to each other. in the second, you get rid of one letter, then solve.

3 0

2 years ago

Which is bigger 57.8% or 0.578 ?

tia_tia [17]

Answer:

57.8%

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

The function f(x)=2/3x-4 was replaced with f(x+k)=2/3x-6 what is the value of k

HACTEHA [7]

Answer:

The given function $f(x)=\frac{2x}{3}-4$ was replaced with $f(x+k)=\frac{2x}{3}-6$ then the value of k is -3

that is k=-3.

Step-by-step explanation:

The given function is $f(x)=\frac{2x}{3}-4\hfill (1)$

and when x=x+k then $f(x+k)=\frac{2x}{3}-6\hfill (2)$

Now to find the value of k:

Put x=x+k in the function f(x)

$f(x)=\frac{2x}{3}-4$

$f(x+k)=\frac{2(x+k)}{3}-4$

$f(x+k)=\frac{2x+2k}{3}-4\hfill (3)$

Now comparing equations (2) and (3) we get

$f(x+k)=\frac{2x+2k}{3}-4=\frac{2x}{3}-6$

$\frac{2x+2k}{3}-4=\frac{2x}{3}-6$

$\frac{2x+2k}{3}-4-\frac{2x}{3}+6=0$

$\frac{2x}{3}+\frac{2k}{3}-4-\frac{2x}{3}+6=0$

$\frac{2k}{3}+2=0$

$\frac{2k}{3}=-2$

$k=\frac{-2\times 3}{2}$

$k=-3$

Therefore k=-3

Therefore the given function $f(x)=\frac{2x}{3}-4$ was replaced with $f(x+k)=\frac{2x}{3}-6$ then the value of k is -3

that is k=-3.

7 0

2 years ago